Problem: $\dfrac{d}{dx}[8\sin(x)-x]=$
Solution: The expression to differentiate includes $\sin(x)$. Remember that the derivative of $\sin(x)$ is $\cos(x)$. Put another way, $\dfrac{d}{dx}[\sin(x)]=\cos(x)$. $\begin{aligned} &\phantom{=}\dfrac{d}{dx}[8\sin(x)-x] \\\\ &=8\dfrac{d}{dx}[\sin(x)]-\dfrac{d}{dx}(x) \\\\ &=8\cdot \cos(x)-1 \\\\ &=8\cos(x)-1 \end{aligned}$ In conclusion, $\dfrac{d}{dx}[8\sin(x)-x]=8\cos(x)-1$